What is the arclength of f(x)=sqrt((x+3)(x/2-1))+5x on x in [6,7]?

1 Answer
May 11, 2017

L~~1.33

Explanation:

The arclength formula is

L=int_a^b sqrt(1+f'(x)) dx

First, find the derivative of the f(x)
d/dx(sqrt((x+3)(x/2-1))+5x)
d/dx(sqrt((x+3)(x/2-1)))+d/dx(5x)
d/dx(sqrt((x+3)(x/2-1)))+5

For the first term, let u=(x+3)(x/2-1)=1/2 x^2+1/2 x -3
(du)/dx =x+1/2

Using u-substitution, find d/dx(sqrt((x+3)(x/2-1)))

d/dx(sqrt(u))=1/(2sqrt(u)) (du)/dx

Replacing these values with x, gives
d/dx(sqrt(u))=(x+1/2)/(2sqrt(1/2 x^2+1/2 x -3))

Plug this result into the arc-length formula above
L=int_a^b sqrt(1+f'(x)) dx
L=int_6^7 sqrt(1+(x+1/2)/(2sqrt(1/2 x^2+1/2 x -3))) dx

This is a very difficult function to integrate, so using numeric methods is recommended. Using Simpson's Rule (with n=10 and Deltax=1//10), you could calculate

L~~(1//10)/3[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+cdots
cdots+4f(x_10)+f(x_11)]~~1.33